Giải thích các bước giải:
Câu 1:
Ta có:
$y'=((x-2)\sqrt{x^2+1})'$
$\to y'=\left(x-2\right)'\sqrt{x^2+1}+\left(\sqrt{x^2+1}\right)'\left(x-2\right)$
$\to y'=1\cdot \sqrt{x^2+1}+\dfrac{x}{\sqrt{x^2+1}}\left(x-2\right)$
$\to y'=\dfrac{2x^2-2x+1}{\sqrt{x^2+1}}$
Câu 2:
Ta có:
$y'=(\dfrac{4}{(2\sqrt{x}+5)^4})'$
$\to y'=4\left(\dfrac{1}{\left(2\sqrt{x}+5\right)^4}\right)'\:$
$\to y'=4\left(\left(2\sqrt{x}+5\right)^{-4}\right)'\:$
$\to y'=4\cdot (\left(2\sqrt{x}+5\right)'\cdot (-4)\cdot \left(2\sqrt{x}+5\right)^{-4-1}$
$\to y'=4\cdot \dfrac{1}{\sqrt{x}}\cdot (-4)\cdot \left(2\sqrt{x}+5\right)^{-5}$
$\to y'=-\dfrac{16}{\sqrt{x}\left(2\sqrt{x}+5\right)^5}$
